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3 years ago

Consider a horse pulling a buggy. Is the

Physics

1 answer:

Degger [83]3 years ago

8 0

Yes, it's true.

But 2nd Newton Law always come to play when the horse is to move forward because obviously the forces interact antagonistically and mass has to be accounted for.

That's what I think. Hope it's right, all the best.

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Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled wi

Korvikt [17]

Complete Question:

Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true?

Object T has a greater density than object B.
Object B has a greater density than object T.
Both objects have the same density.

Answer:

Object B has a greater density than object T

Explanation:

Any object partially or completely submerged in a liquid, experiments an upward force, equal to the weight of the volume displaced by the liquid. This force is called the buoyant force, and can be expressed as follows:

Fb = ρl * Vs*g

where ρl is the density of the liquid, and Vs is the submerged volume.

This force must be compared with the weight of the object, which is always downward, and can be expressed as follows:

Fg = ρb* Vb * g

where ρb, is the density of the object, and Vb is the total volume of the object, regardless which portion is submerged.

For object B, as it floats fully submerged, this means that both forces are equal in magnitude:

Fg = Fb⇒ ρb* Vb * g = ρl * Vs*g

As Vb = Vs (the object is fully submerged) this means that ρb =ρl.

For object T, as it floats partially submerged, this means that Fg < Fb:

Fg= ρt* Vt * g < Fb = ρl * Vs*g.

Now, we know that ρb =ρl, so we can replace in the equation above:

ρT* Vt * g < ρb*Vs*g

Simplifying common terms, and replacing Vs by KVt (where K is the fraction of the total volume which is submerged, i.e. K<1), we have:

ρt*Vt < ρb*K*Vt ⇒ ρt / ρb < K < 1 ⇒ ρt < ρb ⇒ ρb > ρt

3 0

3 years ago

Which of these is not a layer of the skin?

lina2011 [118]

4. hyperdermis is not a layer of skin

3 0

3 years ago

Read 2 more answers

Find the frequency of a wave with the wavelength 3.5 m and the speed is 50 m/s. <br>

Andru [333]

Answer:

8.57 Hz

Explanation:

From the question given above, the following data were obtained:

Wavelength (λ) = 3.5 m

Velocity (v) = 30 m/s

Frequency (f) =?

The velocity, wavelength and frequency of a wave are related according to the equation:

Velocity = wavelength × frequency

v = λ × f

With the above formula, we can simply obtain the frequency of the wave as follow:

Wavelength (λ) = 3.5 m

Velocity (v) = 30 m/s

Frequency (f) =?

v = λ × f

30 = 3.5 × f

Divide both side by 3.5

f = 30 / 3.5

f = 8.57 Hz

Thus, the frequency of the wave is 8.57 Hz

7 0

2 years ago

The figure below shows two forces A=15 N and B=5 N acting on an object. What is the

alexandr1967 [171]

Answer:

18.03 N

Explanation:

From the fiqure below,

Using parallelogram law of vector

R² = 15²+5²-2×5×15cos(180-60)

R² = 225+25-150cos120°

R² = 250-150(-0.5)

R² = 250+75

R² = 325

R = √325

R = 18.03 N

Hence the resultant force of the object is 18.03 N

7 0

3 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago